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lbvjy [14]
2 years ago
12

Find the 26th term in the arithmetic sequence: 20, 26, 32, 38, ...

Mathematics
1 answer:
Dafna11 [192]2 years ago
3 0

Answer:

170

Step-by-step explanation:

20 + (26 - 1)6

= 20 + (25)6

= 20 + 150

= 170

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Jason ate 1/2 of a whole pizza. The next day, he shared the remaining pizza between himself and 2 friends. What fraction of the
suter [353]

Answer:

1/6

Step-by-step explanation:

jason ate 1/2, so only 1/2 of the pizza is left.

1 ÷ 2 = 1/2

he split it equally between himself and 2 friends (3 people)

1/2 ÷ 3

= 1/2 × 1/3

=1/6

8 0
3 years ago
19. Each column corresponds to a probability: 90%, 50%, 25%, 10%, |_%,<br> %
mel-nik [20]

Answer: -9%

Step-by-step explanation:

5 0
2 years ago
A random sample of soil specimens was obtained, and the amount of organic matter (%) in the soil was determined for each specime
MrRissso [65]

Answer:

We conclude that the true average percentage of organic matter in such soil is something other than 3% at 10% significance level.

We conclude that the true average percentage of organic matter in such soil is 3% at 5% significance level.

Step-by-step explanation:

We are given a random sample of soil specimens was obtained, and the amount of organic matter (%) in the soil was determined for each specimen;

1.10, 5.09, 0.97, 1.59, 4.60, 0.32, 0.55, 1.45, 0.14, 4.47, 1.20, 3.50, 5.02, 4.67, 5.22, 2.69, 3.98, 3.17, 3.03, 2.21, 0.69, 4.47, 3.31, 1.17, 0.76, 1.17, 1.57, 2.62, 1.66, 2.05.

Let \mu = <u><em>true average percentage of organic matter</em></u>

So, Null Hypothesis, H_0 : \mu = 3%      {means that the true average percentage of organic matter in such soil is 3%}

Alternate Hypothesis, H_A : \mu \neq 3%      {means that the true average percentage of organic matter in such soil is something other than 3%}

The test statistics that will be used here is <u>One-sample t-test statistics</u> because we don't know about the population standard deviation;

                         T.S.  =  \frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }  ~ t_n_-_1

where, \bar X = sample mean percentage of organic matter = 2.481%

             s = sample standard deviation = 1.616%

            n = sample of soil specimens = 30

So, <u><em>the test statistics</em></u> =  \frac{2.481-3}{\frac{1.616}{\sqrt{30} } }  ~ t_2_9

                                     =  -1.76

The value of t-test statistics is -1.76.

(a) Now, at 10% level of significance the t table gives a critical value of -1.699 and 1.699 at 29 degrees of freedom for the two-tailed test.

Since the value of our test statistics doesn't lie within the range of critical values of t, so we have <u><em>sufficient evidence to reject our null hypothesis</em></u> as it will fall in the rejection region.

Therefore, we conclude that the true average percentage of organic matter in such soil is something other than 3% at 10% significance level.

(b) Now, at 5% level of significance the t table gives a critical value of -2.045 and 2.045 at 29 degrees of freedom for the two-tailed test.

Since the value of our test statistics lies within the range of critical values of t, so we have <u><em>insufficient evidence to reject our null hypothesis</em></u> as it will not fall in the rejection region.

Therefore, we conclude that the true average percentage of organic matter in such soil is 3% at 5% significance level.

8 0
3 years ago
A video game company surveys a random sample of 225 of its best customer and finds that the average gamer spends $606 a year on
victus00 [196]

Answer:

Multi facets

Step-by-step explanation:

First company surveys best customers so they are highest spending

First company standard deviation is high @ $62 so their spends are $606 a year plus or minus $180 (rounded up)

Second company has a Random sample so includes all customers

Second company has larger number of responses

Second company has a range of plus or minus $40

6 0
3 years ago
Compute the odds in favor of obtaining a number divisible by 5 in a single roll of a die.
Georgia [21]

Answer:

1/6 or 16.7% chance

Step-by-step explanation:

A die has 6 sides, and there is only one side with a number divisible by 5, which is the number 5.

So, the probability of rolling a number divisible by 5 is 1/6.

= 1/6 or 0.167

3 0
3 years ago
Read 2 more answers
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